For what value of $m$ does the equation $(x+4)(x+1) = m + 2x$ have exactly one real solution? Express your answer as a common fraction.
Answer: We begin by simplifying the left-hand side of the equation and adding $-m-2x$ to both sides. We get $x^2+3x+(4-m)=0$. For this quadratic to have exactly one real root, the discriminant $b^2-4ac$ must be equal to $0$. Thus, we require $9-4(4-m) = 0$. Solving, we get that $m=\boxed{\frac{7}{4}}$.